POWER SYSTEM ANALYSIS TADP 641 SETTING EXAMPLE FOR OVERCURRENT RELAYS

Transcription

1 POWER SYSTEM ANALYSIS TADP 641 SETTING EXAMPLE FOR OVERCURRENT RELAYS Juan Manuel Gers, PhD

2 Example - Single Line

3 Example 1 - Data Calculate the following: 1. The three phase short circuit levels on busbars 1 and 2 2. The transformation ratios of the CTs associated with breakers 1 to 8, given that the number of primary turns is a multiple of 100 The CT for breaker number 9 is 250/5 3. The settings of the instantaneous elements, and the TAP and DIAL settings of the relays to guarantee a coordinated protection arrangement, allowing a discrimination margin of 0.4 seconds 4. The percentage of the 34.5 kv line protected by the instantaneous element of the overcurrent relay associated with breaker 6 The p.u. impedances are calculated on the following bases: V = 34.5 kv, P = 100 MVA

4 Example 1 - Data The settings of relay 7 are: TAP = 4 A DIAL = 5 Instantaneous = 1100 A All the relays are of the inverse time type with the following characteristics: TAP=1to12A,instepsof1A DIAL = as shown in next figure

5 Calculation of short circuit levels Example V (34,500V ) ZTHEVENIN = = = 6.5 ohms referred to34. 5kV 6 P *10 SC Z transf 1 (34500) = *10 2 = = 7.93ohms, referred to34.5kv ohms, referred to115kv Z transf 2 (34500) = *10 2 = 28.96ohms, referred to34.5kv Z line (34500) = * = 12.93ohms, referred to34.5kv

6 Short circuit levels - Example 1 Z transf1 Z + = Network ( 34.5) = 9.18 ohms, referred to 34.5 kv = ohms, referred to 115 kv Z Network = Ω Ω = Ω, referred to 115 kv

7 Network equivalent - Example 1 Positive Sequence Network (Impedances are referred to 34.5 kv)

8 Nominal currents - Example 1 = A at 34.5 kv I nom.6 = A at 34.5 kv I nom.7 = A at 34.5 kv I nom.8 = A at 34.5 kv = A at 115 kv

9 Short circuit levels - Example 1 = A at 34.5 kv = A at 13.2 kv = A at 34.5 kv

10 Short circuit levels - Example 1 = A at 34.5 kv = A at 34.5 kv = A at 115 kv

11 Example 1 Short circuit printout

12 Selection of current transformers Example 1 The two criteria to be fulfilled are: The CT withstands the nominal current continuously The maximum short circuit current does not saturate the CT. This is verified with the following expression assuming C100 CTs and a burden of 1 Ohm : (Isc 5/X) 100 X (Isc 5/100)

13 Summary of currents and CT ratios Example 1 Breaker No. Pnom (MVA) Inom (A) Isc (A) 5/100 Isc (A) CT Ratio ,2, /5 300/5 200/5 200/5 100/5 200/5 100/5

14 Tap setting - Example 1 I load1,2,3 = A; TAP 1,2,3 = (1.5)(43.74)(5/100) = 3.28; TAP 1,2,3 = 4 A I load4 = 131,22 A; TAP 4 = (1.5)(131.22)(5/200) = 4.92: TAP 4 = 5 A I load5 = A; TAP 5 = (1.5)(50.20)(5/100) = 3.76 A; TAP 5 = 4 A I load6 = A; TAP 6 = (1.5)(50.20)(5/200) = 1.88 A; TAP 6 = 2 A I load8 = A; TAP 8 = (1.5)(251.02)(5/300) = 6.28 A; TAP 8 = 7 A I load9 = A; TAP 9 = (1.5)(75.31)(5/250) = 2.26 A; TAP 9 = 3 A

15 Instantaneous and DIAL setting - Example 1 Relays 1, 2 and 3 Inst. trip = 9 I nom (1/CTR) = (5/100) = A 20 A I prim. trip = 20 (100/5) = 400 A MULT = 20 (1/4) = 5 times. At 5 times and DIAL 1/2, t=0.11 sec. Relay 4 To discriminate with relay 3 at 400 A requires operation in t4 = = 0.51 sec. MULTa = (400)(5/200)(1/5) = 2 times. At 2 times and t op =0.51 sec., DIAL 1/2 MULTb = (0.86)( )(5/200)(1/5) = 4.63 times. At 4.63 times and DIAL 1/2 tb = 0.13 sec.

16 Typical time/current characteristic t relay = 0.71 se``c t 2 = 0.51 sec 0.40 sec t 1 = 0.11 sec

17 Summary of relay settings Example 1 Relay No. CT ratio Tap DIAL Instantaneous 1,2, /5 200/5 100/5 200/5 200/5 300/5 250/ /2 1/ A 26 A 32 A 27.5 A 17 A

18 Coordination curves with a W CO-11

19 Settings with a numerical relay Consider a numerical relay with the following constants: α = 1.0, β = 13.5 L = 0 Therefore t = [(dial) 13.5]/ (MULT 1) Where: MULT = Fault current (in secondary amps)/tap For setting of the instantaneous element consider ten (10) times the maximum load current

20 Settings with a numerical relay Relays 1, 2 and 3 Iinst. trip = 10 Inom (1/CTR) = (5/100) = A set at 22 A. Iprim. trip = 22(100/5) = 440 A. MULT = 22/4 = 5.5 times with dial = 0.05, t = ( )/(5.5 1) = 0.15 s. To coordinate with relays 1,2 and 3 at 440 A, t 4a = = 0.55 s.

21 Settings with a numerical relay Relay 4 MULT 4a = (440)(5/200)(1/5) = 2.2 times. At 2.2 times, and t 4a = 0.55 s, dial = 0.55 (2.2 1)/13.5 = The operating time for a line-to-line fault is determined by taking 86% of the three-phase fault current. MULT 4b = (0.86)( )(5/200)(1/5) = 4.63 times. t 4b = ( )/(4.63 1) = 0.19 s. This relay has no setting for the instantaneous

22 Example 1- Coordination Curves with NR RELAY SETTINGS R1, R2, R3: Tap=4.0 Amp Time Dial=0.05 Inst=22.0 Amp CT=100/5 Amp R4: Tap=5.00 Amp Time Dial=0.05 Inst=Disable CT=200/5 Amp R5: Tap=4.0 Amp Time Dial=0.2 Inst=26.0 Amp CT=100/5 Amp R6: Tap=2.0 Amp Time Dial=0.4 Inst=32.0 Amp CT=200/5 Amp R7: Tap=4.0 Amp Time Dial=0.3 Inst=27.5 Amp CT=200/5 Amp R8: Tap=7.0 Amp Time Dial=0.1 Inst=Disable CT=300/5 Amp R9: Tap=3.0 Amp Time Dial=0.2 Inst=17.0 Amp CT=250/5 Amp THERMAL LIMITS TR1, TR2: Power Transformer L1: Feeder Cable

23 Coverage of the instantaneous unit Example 1 Percentage of 34.5 kv line protected by the instantaneous element of the overcurrent relay associated with breaker 6. Given that: % = K s (1-K i )-1/K i where K i =I sc pickup /I sc end = 1280/ = and K s =Z source /Z element From the computer listing: Z f =V 2 /P = / = 6.5 and, Ks = 6.50/12.92 = 0.50 Therefore: % = 0.50 x ((( )+1)/1.248) = 0.70 so that the instantaneous element covers 70% of the line

24 Minimum short circuit levels When the time delay unit has been set using maximum fault levels, it is necessary to check that the relays will operate at the minimum fault levels and in the correct sequence. For this, it is sufficient to verify that the TAP multiplier under these conditions is greater than 1.5.

25 Questions?